Calculate the work done in joules when 1.0 mole of water vaporizes at 1.0 atm and 100°C. Assume that the volume of liquid water is negligible compared with that of steam at 100°C and ideal gas behavior.

The amount of work done W = - 30.6 J

Explanation:

Pressure = 1 atm

n = 1 mole

T = 100 °c = 373 K

From ideal gas equation

P V = n R T

Put all the values in above equation we get

1 × V = 1 × 0.08206 ×373

V = 30.6 L

Work done is calculated by

W = - P ΔV

W = - 1 × 30.6

W = - 30.6 J

Therefore the amount of work done W = - 30.6 J

The amount of work done W = -3102 J

Explanation:

Pressure = 1 atm

n = 1 mole

T = 100 °c = 373 K

From ideal gas equation

P V = n R T

Put all the values in above equation we get

1 × V = 1 × 0.08206 ×373

V = 30.6 L

Work done is calculated by

W = - P ΔV

W = - 1 atm × 30.6 L

W = - 30.6 atm.L  = -3102 J

Therefore the amount of work done W = -3102 J

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just finished the assignment

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